Conservation of Angular Momentum 8.01 W10D2 Young and Freedman: 10.5-10.6 - PowerPoint PPT Presentation

About This Presentation
Title:

Conservation of Angular Momentum 8.01 W10D2 Young and Freedman: 10.5-10.6

Description:

Conservation of Angular Momentum 8.01 W10D2 Young and Freedman: 10.5-10.6 – PowerPoint PPT presentation

Number of Views:163
Avg rating:3.0/5.0
Slides: 28
Provided by: padour
Learn more at: http://web.mit.edu
Category:

less

Transcript and Presenter's Notes

Title: Conservation of Angular Momentum 8.01 W10D2 Young and Freedman: 10.5-10.6


1
Conservation of Angular Momentum8.01W10D2You
ng and Freedman 10.5-10.6
2
Announcements
Math Review Night Next Tuesday from 9-11 pm Pset
10 Due Nov 15 at 9 pm No Class Friday Nov
11 Exam 3 Tuesday Nov 22 730-930 pm W011D1
Rolling without Slipping, Translation and
Rotation Reading Assignment Young and Freedman
10.3-4
3
Time Derivative of Angular Momentum for a Point
Particle
  • Angular momentum of a point particle about S
  • Time derivative of the angular momentum about S
  • Product rule
  • Key Fact
  • Result

4
Torque and the Time Derivative of Angular
Momentum Point Particle
  • Torque about a point S is equal to the time
    derivative of the angular momentum about S .

5
Conservation of Angular Momentum
  • Rotational dynamics
  • No torques
  • Change in Angular momentum is zero
  • Angular Momentum is conserved

6
Concept Question Change in Angular Momentum
  • A person spins a tennis ball on a string in
    a horizontal circle with velocity (so that
    the axis of rotation is vertical). At the point
    indicated below, the ball is given a sharp blow
    (force ) in the forward direction. This
    causes a change in angular momentum in the
  • direction
  • direction
  • direction

7
Concept Question Change in Angular Momentum
  • Answer 3. The torque about the center of the
    circle points in the positive -direction.
    The change in the angular momentum about the
    center of the circle is proportional to the
    angular impulse about the center of the circle
    which is in the direction of the torque

8
Table Problem Cross-section for Meteor Shower
  • A meteor of mass m is approaching earth as
    shown on the sketch. The distance h on the sketch
    below is called the impact parameter. The radius
    of the earth is Re. The mass of the earth is me.
    Suppose the meteor has an initial speed of v0.
    Assume that the meteor started very far away from
    the earth. Suppose the meteor just grazes the
    earth. You may ignore all other gravitational
    forces except the earth. Find the impact
    parameter h and the cross section ph2 .

9
Angular Momentum for System of Particles
  • Treat each particle separately
  • Angular momentum for system about S

10
Conservation of Angular Momentum for System of
Particles
  • Assumption all internal torques cancel in pairs
  • Rotational dynamics
  • No external torques
  • Change in Angular momentum is zero
  • Angular Momentum is conserved

11
Angular Impulse and Change in Angular Momentum
  • Angular impulse
  • Change in angular momentum
  • Rotational dynamics

12
Kinetic Energy of Cylindrically Symmetric Body
  • A cylindrically symmetric rigid body with moment
    of inertia Iz rotating about its
  • symmetry axis at a constant angular velocity
  • Angular Momentum
  • Kinetic energy

13
Concept Question Figure Skater
  • A figure skater stands on one spot on the ice
    (assumed frictionless) and spins around with her
    arms extended. When she pulls in her arms, she
    reduces her rotational moment of inertia and her
    angular speed increases. Assume that her angular
    momentum is constant. Compared to her initial
    rotational kinetic energy, her rotational kinetic
    energy after she has pulled in her arms must be
  • the same.
  • larger.
  • smaller.
  • not enough information is given to decide.

14
Concept Question Figure Skater
  • Answer 2. Call the rotation axis the z-axis.
    When she pulls her arms in there are no external
    torques in the z-direction so her z-component of
    her angular momentum about the axis of rotation
    is constant. The magnitude of her angular
    momentum about the rotation axis passing through
    the center of the stool is L I?. Her kinetic
    energy is
  • Since L is constant and her moment of inertia
    decreases when she pulls her arms, her kinetic
    energy must increase.

15
Constants of the Motion
  • When are the quantities, angular momentum
    about a point S, energy, and momentum constant
    for a system?
  • No external torques about point S angular
    momentum about S is constant
  • No external work mechanical energy constant
  • No external forces momentum constant

16
Rotational and Translational Comparison
Quantity Rotation Translation
Force
Torque
Kinetic Energy
Momentum
Angular Momentum
Kinetic Energy
17
Demo Rotating on a Chair
  • A person holding dumbbells in his/her arms spins
    in a rotating stool. When he/she pulls the
    dumbbells inward, the moment of inertia changes
    and he/she spins faster.

18
Concept Question Twirling Person
  • A woman, holding dumbbells in her arms, spins
    on a rotating stool. When she pulls the dumbbells
    inward, her moment of inertia about the vertical
    axis passing through her center of mass changes
    and she spins faster. The magnitude of the
    angular momentum about that axis is
  • the same.
  • larger.
  • smaller.
  • not enough information is given to decide.

19
Concept Question Twirling Person
  • Answer 1. Call the rotation axis the z-axis.
    Because we assumed that there are no external
    torques in the z-direction acting on the woman
    then the z-component of her angular momentum
    about the axis of rotation is constant. (Note if
    we approximate the figure skater as a symmetric
    body about the z-axis than there are no
    non-z-components of the angular momentum about a
    point lying along the rotation axis.)

20
Demo Train B134
  • (1) At first the train is started without the
    track moving. The train and the track both move,
    one opposite the other.
  • (2) Then the track is held fixed and the train
    is started. When the track is let go it does not
    revolve until the train is stopped. Then the
    track moves in the direction the train was
    moving.
  • (3) Next try holding the train in place until
    the track comes up to normal speed (Its being
    driven by the train). When you let go the train
    remains in its stationary position while the
    track revolves. You shut the power off to the
    train and the train goes backwards. Put the power
    on and the train remains stationary.

A small gauge HO train is placed on a circular
track that is free to rotate.
21
Table Problem Experiment 6
A steel washer, is mounted on the shaft of a
small motor. The moment of inertia of the motor
and washer is Ir. Assume that the frictional
torque on the axle is independent of angular
speed. The washer is set into motion. When it
reaches an initial angular velocity ?0, at t 0,
the power to the motor is shut off, and the
washer slows down during the time interval ?t1
ta until it reaches an angular velocity of ?a at
time ta. At that instant, a second steel washer
with a moment of inertia Iw is dropped on top of
the first washer. The collision takes place over
a time ?tcol tb - ta.
  1. What is the angular deceleration a1 during the
    interval ?t1 ta?
  2. What is the angular impulse due to the frictional
    torque on the axle during the collision?

c) What is the angular velocity of the two
washers immediately after the collision is
finished?
22
Experiment 6 Conservation of Angular Momentum
23
AppendixAngular Momentum and Torque for a
System of Particles
24
Angular Momentum and Torque for a System of
Particles
  • Change in total angular momentum about a point S
    equals the total torque about the point S

25
Internal and External Torques
  • Total external torque
  • Total internal torque
  • Total torque about S

26
Internal Torques
  • Internal forces cancel in pairs
  • Does the same statement hold about pairs of
    internal torques?
  • By the Third Law this sum becomes

The vector points from the
jth element to the ith element.
27
Central Forces Internal Torques Cancel in Pairs
  • Assume all internal forces between a pair of
    particles are directed along the line joining the
    two particles then
  • With this assumption, the total torque is just
    due to the external forces
  • No isolated system has been encountered such
    that the angular momentum is not constant.
Write a Comment
User Comments (0)
About PowerShow.com